\documentclass[12pt]{article}
\usepackage{calc,fancyhdr}
\usepackage[hmargin=.5in,vmargin=.75in,
            footskip=.25in,headsep=.5in-\headheight]{geometry}

% Template solution to Exercise 5.

% Common formatting macros for CSC165.
\usepackage{amsmath,amssymb}
\input{macros-165}

% Page layout: stretch text to fill up page.
\flushbottom

% Headings.
\pagestyle{fancy}
\let\headrule\empty
\let\footrule\empty
\lhead{CSC\,165\,H1S}
\chead{Homework Exercise \#\,5}
\rhead{Winter 2014}

\begin{document}

\begin{large}
  \noindent
  Partner (1): Andrea Hunter \hfill CDF login name: c3hunter\\[0.5cm]
  Partner (A): David Keon    \hfill CDF login name: c3keonda
\end{large}

\medskip

\noindent
\rule{\textwidth}{.5pt}

\subsection*{Topic: Proofs I}

\medskip

\begin{enumerate}
 
       
    % place solution to question 1 below

    \item  fill it in
    
    % place solution to question 2 below

    \item  fill it in

    % place solution to question 3 below

    \item fill it in

    % place solution to question 4 below

    \item fill it in

\end{enumerate}

\noindent
And here is a sample that demonstrates how you can present your
proof in \LaTeX.  

\medskip
\noindent
Let $a > 0$ be a real number and consider the following claim:
\begin{quote}
If $a < 1$ then $\sqrt{a} > a.$
\end{quote}
We can express this in the notation of symbolic logic as:
$$\forall\ a \in \R^{+}, [(a < 1) \implies (\sqrt{a} > a)].$$
Here $\R^{+}$ stands for the set of positive real numbers.

\noindent
Here is a proof of the statement that makes use of the macros
\verb|pindent|, \verb|assumption| and \verb|just| that are defined in
the \verb|macros-165| file.

\noindent
{\bf Proof:}

\begin{pindent}
    \begin{assumption}{$a \in \R^{+}.$}
        Then $a > 0.$ 
        \begin{assumption}{$\sqrt{a} \le a$.}
            Then $\sqrt{a}\cdot\sqrt{a} \le a\cdot a.$
                \just{multiplying by smaller, positive value on left
                      maintains the inequality}\\
            Then $a \le a^2.$\\
            Then $0 \le a^2-a.$\\
            Then $0 \le a(a-1).$\\
            Then $0 \le a-1.$ \just{Since $a > 0.$}\\
            Then $1 \le a.$ \\
            Then $a \ge 1.$
        \end{assumption}{Then $(\sqrt{a} \le a) \implies (a \ge 1).$}\\
        Then $(a < 1) \implies (\sqrt{a} > a).$ \just{By contrapositive.}
    \end{assumption}{Then $\forall\ a \in \R^{+}, (a < 1) \implies (\sqrt{a} > a).$}
\end{pindent}
\medskip

\noindent
\rule{\textwidth}{.5pt}

\end{document}
